Question

我有一个表单中的输入，它发送一个ajax请求（带有serializeArray()），并根据这些检查的输入，查询得到不同的数据（如过滤器）。但是我在处理这些动态条件时遇到了问题，如果它只有1（AND）它正在工作，但由于某种原因它会超过1它会抛出一个错误（显然一切都很好）：

MySQLi failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 't2.type = 2
 OR t2.type = 3

                                   LIMIT ?,' at line 5<br><br />

问题必须是如何创建变量或如何在prepare()语句中打印：

$filter_type = isset( $_POST['type'] ) ?  filter_input( INPUT_POST , 'type', FILTER_SANITIZE_NUMBER_INT , FILTER_REQUIRE_ARRAY ) : NULL;

if ($filter_type != NULL) {
    foreach ($filter_type as $type=>$value) {
        if ($type === 0) {
            $type_cond = "\nAND t2.type = " . $value; // (int) instead of filter_input()?
        } else {
            $type_cond .= $type_cond . "\nOR t2.type = " . $value;
        }
    }
}

// I manually set $table, based on post (but not directly)
if ($stmt = $mysqli->prepare(' SELECT t1.id, t2.*
                               FROM ' . $table . ' t1
                               INNER JOIN property t2 ON t2.id = t1.id
                               WHERE t2.use = ?
                               ' . $type_cond
                                 . $filter_status
                                 . $filter_bhk
                                 . $filter_city
                                 . $filter_zone . '
                               LIMIT ?, ?')) {

问题出在哪里？

编辑：~~预期的~~实际查询，在选中2个复选框后将是$filter_type = array(2, 3)所以：

SELECT t1.id, t2.*
FROM for_sale t1
INNER JOIN property t2 ON t2.id = t1.id
WHERE t2.use = 1AND t2.type = 2
OR t2.type = 3
-- $filter_status = NULL; but not printing
-- $filter_bhk = NULL; but not printing
-- $filter_city = NULL; but not printing
-- $filter_zone = NULL; but not printing
LIMIT ?, ?

由@MarcB修改（在评论中）。

$filter_type = "\nAND t2.ty...";

mysql选择的动态条件

0 个答案: